package com.summli.basic.class07;

/**
 * 二叉树结构如下定义：
 * Class Node {
 * 	V value;
 * 	Node left;
 * 	Node right;
 * 	Node parent;
 * }
 * 给你二叉树中的某个节点，返回该节点的后继节点
 */
public class Code07_SuccessorNode {
    public static class Node {
        public int value;
        public Node left;
        public Node right;
        public Node parent;

        public Node(int data) {
            this.value = data;
        }
    }
    //传统的方法，可以找到head节点然后中序遍历，可以找到后续节点
    //时间复杂度可以达到O(K)
    // 思路 如果当前节点有右节点，那么后续节点就是右子树的最左节点
    // 如果当前节点没有右节点，那么后续节点就是当前节点的父节点（第一个当前节点是父节点的左孩子，那么这个父节点就是当前节点的后续节点）
    public static Node getSuccessorNode(Node node) {
        if(node == null){
            return null;
        }
        if(node.right != null){
            return getleftNode(node.right);
        }else{
            Node parent = node.parent;
            while(parent != null){
                if(parent.left == node){
                    return parent;
                }else{
                    node = parent;
                    parent = parent.parent;
                }
            }
        }
        return null;
    }

    private static Node getleftNode(Node node) {
        while(node.left!= null){
            node = node.left;
        }
        return node;
    }

    public static void main(String[] args) {
        Node head = new Node(6);
        head.parent = null;
        head.left = new Node(3);
        head.left.parent = head;
        head.left.left = new Node(1);
        head.left.left.parent = head.left;
        head.left.left.right = new Node(2);
        head.left.left.right.parent = head.left.left;
        head.left.right = new Node(4);
        head.left.right.parent = head.left;
        head.left.right.right = new Node(5);
        head.left.right.right.parent = head.left.right;
        head.right = new Node(9);
        head.right.parent = head;
        head.right.left = new Node(8);
        head.right.left.parent = head.right;
        head.right.left.left = new Node(7);
        head.right.left.left.parent = head.right.left;
        head.right.right = new Node(10);
        head.right.right.parent = head.right;

        Node test = head.left.left;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.left.left.right;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.left;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.left.right;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.left.right.right;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.right.left.left;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.right.left;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.right;
        System.out.println(test.value + " next: " + getSuccessorNode(test).value);
        test = head.right.right; // 10's next is null
        System.out.println(test.value + " next: " + getSuccessorNode(test));
    }
}
